8. Properties of Curves

i. Frenet Frames (Optional)

1. Frenet Frames & Derivatives of TNB

\(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) make up what is called the Frenet Frame. This is a right handed set of \(3\) mutually orthogonal unit vectors that moves along the curve, as opposed to the \(\hat{\imath},\hat{\jmath},\hat{k}\) unit vectors which make up the fixed standard reference frame for \(\mathbb{R}^3\).

Algebraic Properties

The Frenet Frame can be used much like the standard frame. As a right handed set of \(3\) orthogonal unit vectors, \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) satisfy: \[\begin{aligned} \hat{T}\cdot\hat{T}&=1 \qquad &\hat{N}\cdot\hat{N}&=1 \qquad &\hat{B}\cdot\hat{B}&=1 \\ \hat{T}\cdot\hat{N}&=0 \qquad &\hat{N}\cdot\hat{B}&=0 \qquad &\hat{B}\cdot\hat{T}&=0 \qquad (*) \\ \hat{T}\times\hat{N}&=\hat{B} \qquad &\hat{N}\times\hat{B}&=\hat{T} \qquad &\hat{B}\times\hat{T}&=\hat{N} \end{aligned}\]

Suppose \(\vec u=u_T\hat T + u_N\hat N + u_B\hat B\) and \(\vec v=v_T\hat T + v_N\hat N + v_B\hat B\).

1. Use the distributive rule, the anti-commutative rule and conditions (*) to compute \(\vec u\times\vec v\).

   \( \vec u\times\vec v =(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B \)

   Using the distributive rule: \[\begin{aligned} \vec u\times\vec v &=(u_T\hat T + u_N\hat N + u_B\hat B)\times(u_T\hat T + u_N\hat N + u_B\hat B) \\ &= u_Tv_T\hat T\times\hat T + u_Tv_N\hat T\times\hat N + u_Tv_B\hat T\times\hat B \\ &\quad + u_Nv_T\hat N\times\hat T + u_Nv_N\hat N\times\hat N + u_Nv_B\hat N\times\hat B \\ &\quad + u_Bv_T\hat B\times\hat T + u_Bv_N\hat B\times\hat N + u_Bv_B\hat B\times\hat B \\ \end{aligned}\] Then using (*) and the anti-commutative rule: \[\begin{aligned} \vec u\times\vec v &= u_Tv_T(\vec 0) + u_Tv_N(\vec B) + u_Tv_B(-\vec N) \\ &\quad + u_Nv_T(-\vec B) + u_Nv_N(\vec 0) + u_Nv_B(\vec T) \\ &\quad + u_Bv_T(\vec N) + u_Bv_N(-\vec T) + u_Bv_B(\vec 0) \\ &=(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B \end{aligned}\]

2. Compute \(\begin{aligned} \begin{vmatrix} \hat T & \hat N & \hat B \\ u_T & u_N & u_B \\ v_T & v_N & v_B \end{vmatrix} \end{aligned}\) by expanding on the first row. What do you notice?

   \(\begin{aligned} \begin{vmatrix} \hat T & \hat N & \hat B \\ u_T & u_N & u_B \\ v_T & v_N & v_B \end{vmatrix} &=(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B \\ \end{aligned}\)
   We notice, this is the same as \(\vec u\times\vec v\).

   \[\begin{aligned} \begin{vmatrix} \hat T & \hat N & \hat B \\ u_T & u_N & u_B \\ v_T & v_N & v_B \end{vmatrix} &=\hat T \begin{vmatrix} u_N & u_B \\ v_N & v_B \end{vmatrix} -\hat N \begin{vmatrix} u_T & u_B \\ v_T & v_B \end{vmatrix} +\hat B \begin{vmatrix} u_T & u_N \\ v_T & v_N \end{vmatrix} \\ &=(u_Nv_B-u_Bv_N)\hat T-(u_Tv_B-u_Bv_T)\hat N+(u_Tv_N-u_Nv_T)\hat B \\ \end{aligned}\] We notice, this is the same as \(\vec u\times\vec v\).

These two exercises show that the dot and cross products can be computed using components relative to \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) in exactly the same way as they are computed using components relative to \(\hat\imath\), \(\hat\jmath\) and \(\hat k\).

Differential Properties

We now turn to the derivative properties of Frenet frames. First notice that a vector field, \(\vec F(t)\), along a curve, \(\vec r(t)\), can be expanded in either the \(\hat\imath\hat\jmath\hat k\)-basis or the \(\hat{T}\hat{N}\hat{B}\)-basis: \[\begin{aligned} \vec F&=F_1\hat\imath + F_2\hat\jmath + F_3\hat k \\ &=F_T\hat T + F_N\hat N + F_B\hat B \end{aligned}\] We want to differentiate \(\vec F\). In the \(\hat\imath\hat\jmath\hat k\) form, since \(\hat\imath\), \(\hat\jmath\) and \(\hat k\) are constant, we only need to differentiate the coefficients: \[ \dfrac{d\vec F}{dt} =\dfrac{dF_1}{dt}\hat\imath + \dfrac{dF_2}{dt}\hat\jmath + \dfrac{dF_3}{dt}\hat k \] However, in the \(\hat{T}\hat{N}\hat{B}\) form, \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) are not constant, and so we also need to differentiate them. By the Product Rule: \[\begin{aligned} \dfrac{d\vec F}{dt} &=\dfrac{dF_T}{dt}\hat T + F_T\dfrac{d\hat T}{dt} \\ &\quad+ \dfrac{dF_N}{dt}\hat N + F_N\dfrac{d\hat N}{dt} \\ &\quad\qquad+ \dfrac{dF_B}{dt}\hat B + F_B\dfrac{d\hat B}{dt} \end{aligned}\] The Frenet formulas tell us the derivatives of \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) (with respect to arclength, \(s\)) written as linear combinations of \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\). The coefficients turn out to be the curvature, \(\kappa\), and the torsion, \(\tau\). The derivatives with respect to time, \(t\), are given as a corollary.

The derivatives of \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) satisfy: \[\begin{aligned} \dfrac{d\hat{T}}{ds}&=\qquad \quad \kappa \hat{N} \\ \dfrac{d\hat{N}}{ds}&=-\,\kappa \hat{T} \quad + \quad \tau \hat{B} \\ \dfrac{d\hat{B}}{ds}&=\qquad -\,\tau \hat{N} \end{aligned}\] Proof

Since \(\dfrac{d}{dt}=\dfrac{ds}{dt}\dfrac{d}{ds}=|\vec v|\dfrac{d}{ds}\) we conclude:

The derivatives of \(\hat{T}\), \(\hat{N}\) and \(\hat{B}\) satisfy: \[\begin{aligned} \dfrac{d\hat{T}}{dt}&=\qquad \quad |\vec v|\kappa \hat{N} \\ \dfrac{d\hat{N}}{dt}&=-\,|\vec v|\kappa \hat{T} \quad + \quad |\vec v|\tau \hat{B} \\ \dfrac{d\hat{B}}{dt}&=\qquad -\,|\vec v|\tau \hat{N} \end{aligned}\]

Given \(\vec F = 2t\hat{T} + (1+2t^2) \hat{N} - 4t^3 \hat{B}\), find \( \dfrac{d \vec F}{dt} \) on the twisted cubic \(\vec{r}=\left\langle t,t^2,\dfrac{2}{3}t^{3}\right\rangle\).

First, we differentiate with respect to \(t\) using the Product Rule: \[\begin{aligned} \dfrac{ d \vec F}{dt} = 2\hat{T} + 2t \dfrac{d \hat{T}}{dt} + 4t \hat{N} + (1+2t^2) \dfrac{d \hat{N}}{dt} - 12t^2 \hat{B}\ - 4t^3 \dfrac{d \hat{B}}{dt} \end{aligned}\] Recall that for this twisted cubic \[\begin{aligned} |\vec{v}| &= 1+2t^2 \\ \kappa &= \dfrac{2}{(1+2t^2)^2}\\ \tau &= \dfrac{2}{(1+2t^2)^2} \end{aligned}\] So from the Frenet Formulas: \[\begin{aligned} \dfrac{d\hat{T}}{dt}&= \dfrac{2}{1+2t^2} \hat{N} \\ \dfrac{d\hat{N}}{dt}&= -\dfrac{2}{1+2t^2} \hat{T} + \dfrac{2}{1+2t^2} \hat{B} \\ \dfrac{d\hat{B}}{dt}&= -\dfrac{2}{1+2t^2} \hat{N} \end{aligned}\] We substitute these values into our derivative: \[\begin{aligned} \dfrac{d\vec F}{dt} &= 2\hat{T} + 2t \dfrac{2}{1+2t^2} \hat{N} + 4t \hat{N} \\ &\quad+(1+2t^2) \left(-\,\dfrac{2}{1+2t^2} \hat{T} + \dfrac{2}{1+2t^2} \hat{B}\right) \\ &\quad-12t^2 \hat{B} - 4t^3 \left(-\,\dfrac{2}{1+2t^2} \hat{N}\right) \\ &= 2\hat{T} + \dfrac{4t}{1+2t^2} \hat{N} + 4t \hat{N} -2 \hat{T} + 2 \hat{B} - 12t^2 \hat{B}\ + \dfrac{8t^3}{1+2t^2} \hat{N} \\ &= \left(\dfrac{4t + 8t^3}{1+2t^2} + 4t\right) \hat{N}+ (2 - 12t^2) \hat{B} \\ &= 8t\hat{N} + (2 - 12t^2) \hat{B} \end{aligned}\]